card trick without the trick...
The question could've been a bit more specific by stating "i then arrange the other four cards in a special way" as "i then 'order' the other four cards in a special way".
I figured that If i'm allowed to encode the card by "arranging" the four cards, I can keeo each card in either vertical or horizontal positions. This would allow me to encode 2^4 = 16 values. Ofcourse this is not enough so we move on a higher base system. I now keep a card horizontal or vertical in addition to keeping it above or below a reference line(or some such modification that allows us to use a higher base system e.g. keep the carc vertical, horizontal or tilted). This would encode 4^4=256 more than enough for the 52 value that we have...
Ofcourse its a minor issue... But I got the solution wrong :(
Anyways, thanks for the problems ... I'm addicted to puzzle solving.
~ Amol
Amol Deshmukh
Monday, December 2, 2002
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